and beginning with the units' column, we add thus, -4, 13, 18, 25, 32, 40 units, equal to 4 tens and 0 units. Writing 0 in the units' place in the answer, we add the 4 tens with the tens of the tens' column, thus, - 4, 10, 13, 15, 17, 19, 28 tens, equal to 2 hundred and 8 tens. Writing the 8 tens, we add the 2 hundred with the hundreds of the column of hundreds, thus, 2, 7, 10, 11, 15, 19, 26 hundreds, equal to 2 thousands and 6 hundreds. Writing the 6 hundreds, we add the 2 thousands with the thousands of the column of thousands, thus, -2, 3, 10, 13 thousands, equal to 1 ten-thousand and 3 thousands, which we write.

Having added all the denominations of each number, we must have the sum of the numbers, and hence the answer, which is 13680 dollars.

Example 2. What is the sum of 4 cwt. 2 qr. 19 lb. 6 oz. 15 dr.; 2 cwt. 1 qr. 21 lb. 11 oz.; 3 cwt. 3 qr. 27 lb. 13 oz. 11 dr.; and 5 cwt. 2 qr. 26 lb. 14 oz. 3 dr.?

Explanation. Commencing with the drams' column, we add thus, -3, 14, 29 drams, equal to 1 oz. 13 dr. Writing the 13 drams in the answer, and adding the 1 ounce with the ounces, we have 1, 15, 28, 39, 45 ounces, equal to 2 lb. 13 oz. Writing the 13 oz., and adding the 2 lb. with the pounds, we have 2, 28, 55, 76, 95 pounds, equal to 3 qr. 11 lb. Writing the 11 lb., and adding the 3 qr. with the quarters, we have 3, 5, 8, 9, 11 quarters, equal to 2 cwt. 3 qr. Writing the 3 qr., and adding the 2 cwt. with the hundred weight, we have 7, 10, 12, 16 hundred weight, which we write. As all the denominations have now been added, we must have obtained the sum, which is 16 cwt. 3 qr. 11 lb. 13 oz. 13 dr.

C. We can test the correctness of the work in many ways, a few of which we will mention. One very good method is to go over the work carefully a second time in the same manner as at first. Another is to begin at the top to add instead of begin

ning at the bottom. This, by presenting the figures in a different order, renders it improbable that any mistake which may have been made in the first work will be repeated. Another method is to begin at the left hand to add, writing the sum of each denomination separately, as below, and then add these sums together.

so that we may suppose that the work is correct.

which being the same as the one before obtained, we may suppose that the work is correct.

Another method is to begin at the left, and proceed as follows. The sum of the thousands' column, in example 1, is 11 thousand; but as in the answer first obtained there are 13 thousands, two thousands must have been obtained from the sum of

1,

13.

"dr.

66

11, 13,

the lower denominations, or else the answer is wrong. But 2 thousands equal 20 hundreds, and there are 6 hundreds written in the hundreds' place; hence we should expect to find 26 hundreds in the sum of the hundreds' column. Since there are but 24 hundreds in that sum, 2 hundreds must have been obtained from the lower denominations, or else the work is incorrect. But 2 hundreds equal 20 tens, and 8 tens are written in the tens' place; hence we should expect to find 28 tens in the sum of the tens' column. Since there are but 24 tens in that sum, 4 tens must have come from the column of units; 4 tens equal 40 units, and 0 being written in the units' place, we should expect to find 40 units in the sum of the units' column, and as that sum is 40, the work may be presumed to be right.

but as

In example 2, the sum of the cwt. column is 14 cwt., in the answer first obtained there are 16 cwt., 2 cwt. or 8 qr. must have been obtained from the sum of the lower denominations, or else the work is wrong. 8 qr. + 3 qr. 11 qr., which we must account for. As the sum of the quarters' column is only 8 qr., 3 qr. or 84 lb. must have been brought from the lower denominations. 84 lb. + 11 lb. 95 lb., which we must account for. As the sum of the pounds' column is only 93 lb., 2 lb. or 32 oz. must have been brought from the lower denominations, or else the work is wrong. 32 oz. 13 oz. = 45 oz., which we must account for. As the sum of the ounces' column is only 44 oz., 1 oz. or 16 dr. must have been brought from the drams, or else the work is wrong. 16 dr. + 13 dr. = 29 dr., which we must account for; and as the sum of the drams is just 29 drams, we may infer that our work is correct.

If, by any of these methods, we obtain a different result from the one we first obtained, we may be sure there is an error in one or both operations, and should examine both carefully to find the error or errors.

Some method of proof should always be resorted to, until the pupil acquires sufficient skill to be sure of the accuracy of his work without.

D. When numbers used to express any value, express it in terms of a single denomination, as in pence, in shillings, in dollars, or in bushels, they are called simple numbers; but when they express it in terms of several different denominations they are called compound numbers.

The numbers in example 1 are simple, but those in example 2 are compound.

The principles involved in Simple Addition (the addition of

simple numbers) are exactly the same as tnose involved in Compound Addition (the addition of compound numbers).

IMPORTANT NOTE. The method given in this section illustrates the similarity of simple and compound addition. As soon as understood, it should be extended so as to include the addition of two, three, or more columns at a time, (See Section 6th, Letters C. and G.) and in compound addition the reductions should be made on the principles illustrated in Section 7th, Letters R. and S.

The pupil should master all the methods, and then, in each particular case, choose the one which seems best adapted to that case. One process may be used to prove the correctness of a result obtained by some other.

Add the numbers in the following examples:

21. How many are 831 + 427 +1058 +69486 +37 + 499 +54638?

* Obtaining the answer to the 12th example in the usual method, we shall find it to be 24 rd. 5 yd. 2 ft. 11 in. This is correct, but it is not in the best form, for although there are not units enough expressed of any denomination to make one of the next higher, it equals 25 rd. 1 ft. 5 in. Show the truth of this statement, and show also why the answer does not at first appear in the best form.

The answer to the 13th example will take the form at first of 35 rd. 4 yd. 2 ft. 11 in., which should be changed, for the sake of simplicity, to 35 rd. 5 yd. 1 ft. 5 in. Show the equality of the two expressions, and the method by which the reduction can be made.